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I have some questions following from a related question I did Interpretation of fixed point's paths as functions defined on $\mathbb{S}^{1}$

Now let $\phi : \pi_{1}(X,x_{0}) \longmapsto [S^{1},X]$, where $[S^{1},X]$ the free homotopic classes of continuos function from $S^{1} \longmapsto X$. The maps $\phi$ send the class of $[\gamma]$ to the class of $[\tilde{\gamma}]$ by the natural identification of a loop as a function defined on $S^{1}$.

The properties of $\phi$ I'd like to prove are :

$\textbf{(1)}$ $\phi$ is an homomorphism.

$\textbf{(2)}$ $\phi$ is surjective $\textit{if}$ $X$ $\textit{is path connected}$ .

$\textbf{(3)}$ $\phi([\gamma]) = \phi([\gamma']) \iff [\gamma]$ is conjugated to $[\gamma']$ in $\pi_{1}(X,x_{0})$.

I was able to prove that is an homomorpshim, and as far as it concerns the surjectivity I manage to tell that if $\gamma$ is a path from $x_{0}$ to $x_{1}$, $\alpha$ from $x_{1}$ to $x_{2}$ and $\beta$ from $x_{2}$ to $x_{0}$ then the concatenion of those three (in any order that creates a loop) are homotopic with fixed points loops (the idea I had in mind was to compose with a rotation) but I stuck since I couldn't find the homotopy explicitly which would give me all the pieces to make a complete proof.

About the third I found problem in the $[\Rightarrow]$ implication, since if $\phi([\gamma]) = \phi([\gamma'])$ $\hspace{0.1cm} \exists H(t,s) = \begin{cases}H(t,0) = \gamma(t) & \forall t \in [0,1]\\ H(t,1) = \gamma'(t) & \forall t \in [0,1] \\ H(0,s) = H(1,s) & \forall s \in [0,1] \end{cases}$. Now let $h$ be a loop in $x_{0}$ $h(s) = H(0,s)= H(1,s)$, then I'd like to explicit an homotopy $F(t,s)$ with fixed endpoints between $\gamma$ and $h \star \gamma' \star \bar{h}$, which I was unable to do.

Any help would be appreciated. There is a lot references and notes on this subject even though I can't find any explicit solution, in particular for $\textbf{(3)}$.

Se some references here : Conjugacy classes in the fundamental group, Conjugation in fundamental group,

jacopoburelli
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    Your formulation of the problem has some errors. In particular, the set $[S^1,X]$ is not a group, so statement (1) as written makes no sense. – Lee Mosher Aug 29 '20 at 21:19
  • $\phi$ is not surjective if $X$ is not path-connected. – Paul Frost Aug 29 '20 at 22:12
  • I edit the error $\textbf{(2)}$ in the text, @LeeMosher I see, what the map should be defined like ? – jacopoburelli Aug 30 '20 at 06:01
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    The definition of the map $\phi$ is fine. The point is that there is no group structure on the set $[S^1,X]$. Therefore it makes no sense to even assert (let alone to prove) that the map $\phi$ is a homomorphism, because its codomain $[S^1,X]$ is not a group. – Lee Mosher Aug 30 '20 at 14:21

1 Answers1

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I assume $X$ is path connected, otherwise the statements are not valid.

  1. As written in the comments, this statement makes no sense since in general $[S^1,X]$ doesn't have a group structure (unless e.g. $X$ is a topological group).

  2. Fix a point $s\in S^1$ and let $f:S^1\to X$ be arbitrary. Because of path connectivity there's a path $p:x_0\leadsto f(s)$, and $f$ is homotopic to the path composition $pf p^{-1}$ which is a loop on $x_0$. (Let the $t$th layer use only $p|_{[1-t,\,t]}$.)

  3. If $H$ is a free homotopy between loops $\gamma$ and $\gamma'$ as you write, then consider the loop $\sigma:=t\mapsto H(0,t)$.
    Now imagine the continuous transformation of the square $[0,1]^2$ that shrinks its top to the third, lifts its left and right side to append them to the top, while fixing the bottom and opening up the two lower corners to become the new left and right sides.
    Composing this with $H$ gives a fixed homotopy between $\gamma$ and $\sigma\gamma'\sigma^{-1}$.

Alternatively, you can explicitly write up a specific version of the above idea for 3. where the $t$th level of the new homotopy is the composition of paths $\sigma|_{[0,t]}\, H(-,t)\,{\sigma|_{[0,t]}}^{-1}$, parametrized so that the parts $\sigma|_{[0,t]}$ on the left and its inverse on the right both take $t/3$ time (and thus the middle part is reparametrized to take $1-\frac{2t}3$ time from the whole domain interval $[0,1]$.
Can you write up a formula for this?

Berci
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  • I don't see how the continuos trasformation does look like, can you please be more specific about "opening up the two lower corners to become the new left and right sides" ? – jacopoburelli Aug 30 '20 at 06:07
  • Specifically, the left side ${0}\times [0,1]$ is mapped constantly to the lower left corner $(0,0)$ and similarly the right side is mapped to the lower right corner. – Berci Aug 30 '20 at 08:43